klastiesym

2022-10-24

Find the Laplace transform of the function $H\left(t-a\right){t}^{n}$. That's what I have done so far:
$L\left\{H\left(t-a\right){t}^{n}\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}H\left(t-a\right){t}^{n}dt={\int }_{a}^{\mathrm{\infty }}{e}^{-st}{t}^{n}dt={\int }_{0}^{\mathrm{\infty }}{e}^{-s\left(t+a\right)}\left(t+a{\right)}^{n}dt$
Could you tell me how I can continue?

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t5an1izqp

Expert

You can obtain directly from the integral
$\mathcal{L}\left\{H\left(t-a\right){t}^{n}\right\}={\int }_{a}^{\mathrm{\infty }}{e}^{-st}{t}^{n}dt=\frac{1}{{s}^{n+1}}{\int }_{as}^{\mathrm{\infty }}{e}^{-z}{z}^{n}dz=\frac{1}{{s}^{n+1}}\mathrm{\Gamma }\left(n+1,as\right)$
recalling the integral expression of incomplete Gamma function for integer n
$\mathrm{\Gamma }\left(n,x\right)={\int }_{x}^{\mathrm{\infty }}{e}^{-z}{z}^{n-1}dz.$
NOTE Another useful methodology may be the following.
Observing that
$H\left(t-a\right){t}^{n}=H\left(t-a\right)\left(t-a+a{\right)}^{n}=H\left(t-a\right)\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\left(t-a{\right)}^{k}{a}^{n-k}$
the Laplace transform becomes
$\mathcal{L}\left\{H\left(t-a\right){t}^{n}\right\}=\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){a}^{n-k}\mathcal{L}\left\{H\left(t-a\right)\left(t-a{\right)}^{k}\right\}$
recalling that $\mathcal{L}\left\{H\left(t-a\right)f\left(t-a\right)\right\}={e}^{-as}F\left(s\right)$. Using $\mathcal{L}\left\{{t}^{n}\right\}=\frac{n!}{{s}^{n+1}}$, we obtain
$\mathcal{L}\left\{H\left(t-a\right){t}^{n}\right\}={e}^{-as}\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){a}^{n-k}\frac{k!}{{s}^{k+1}}=\frac{n!{e}^{-as}}{{s}^{n+1}}\sum _{k=0}^{n}\frac{\left(as{\right)}^{n-k}}{\left(n-k\right)!}=\frac{1}{{s}^{n+1}}n!{e}^{-as}\sum _{\nu =0}^{n}\frac{\left(as{\right)}^{\nu }}{\nu !}$
and recalling that
$\mathrm{\Gamma }\left(n,z\right)=\left(n-1\right)!{e}^{-z}\sum _{\nu =0}^{n-1}\frac{{z}^{\nu }}{\nu !}$
is the incomplete Gamma function for integer n, we obtain finally
$\mathcal{L}\left\{H\left(t-a\right){t}^{n}\right\}=\frac{1}{{s}^{n+1}}\mathrm{\Gamma }\left(n+1,as\right).$

Still Have Questions?

Pellagra3d

Expert

You know that $\mathcal{L}\left\{\mathcal{U}\left(t-a\right)\right\}=\frac{{e}^{-as}}{s},a>0$. And
$\mathcal{L}\left\{{t}^{n}f\left(t\right)\right\}=\left(-1{\right)}^{n}\left(\mathcal{L}\left(f\left(t\right)\right){\right)}^{\left(n\right)}$
Now let think about $\left(\frac{{e}^{-as}}{s}{\right)}^{\left(n\right)}$. Using Maple, I found:
$n=1\to {\left(\frac{{e}^{-as}}{s}\right)}^{\prime }=-\mathrm{exp}\left(-as\right)\frac{as+1}{{s}^{2}}\phantom{\rule{0ex}{0ex}}n=2\to {\left(\frac{{e}^{-as}}{s}\right)}^{″}=\mathrm{exp}\left(-as\right)\frac{{a}^{2}{s}^{2}+2as+2}{{s}^{3}}\phantom{\rule{0ex}{0ex}}n=3\to {\left(\frac{{e}^{-as}}{s}\right)}^{‴}=-\mathrm{exp}\left(-as\right)\frac{{a}^{3}{s}^{3}+3{a}^{2}{s}^{2}+6as+6}{{s}^{4}}$
And so I think the possible pattern could be
$\left(-1{\right)}^{n}\frac{\sum _{k=0}^{n}\frac{n!}{\left(n-k\right)!}{s}^{n-k}{a}^{n-k}}{{s}^{n+1}}$

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